How to solve $\max|x_1|\Sigma_{i=1}^n|x_i|$, with constraint $\Sigma_{i=1}^n|x_i|^2=1,n>2$ Alternatively, that making the upper bound for $|x_1|\Sigma_{i=1}^n|x_i|$ as tight as possible is also welcome. For me, if $\Sigma_{i=1}^n|x_i|=\sqrt{n}$, then $|x_1|=\sqrt{1/n}$. If $|x_1|=1$,then $\Sigma_{i=1}^n|x_i|=1$. In a word, I guess the objective function may be bounded by a constant.
 A: WLOG, $x_i\geq 0$ for every $i$.  From $\sum_{i>1}x_i\leq \sqrt{(n-1)\sum_{i>1}x_i^2}=\sqrt{(n-1)\left(1-x_1^2\right)}$, we have $x_1\sum_{i=1}^nx_i\leq x_1^2+x_1\sqrt{(n-1)\left(1-x_1^2\right)}$.  By AM-GM, $\sqrt{ax_1^2\cdot b\left(1-x_1^2\right)}\leq \frac{b}{2}-x_1^2$, where $0\leq a<b$ satisfy $ab=n-1$ and $b-a=2$.  Note that $a=\sqrt{n}-1$ and $b=\sqrt{n}+1$.  Hence, $x_1\sum_{i=1}^nx_i\leq \frac{b}{2}=\frac{\sqrt{n}+1}{2}$.  This bound is sharp for all positive integers $n$, i.e., by taking $x_1=\sqrt{\frac{b}{a+b}}=\sqrt{\frac{\sqrt{n}+1}{2\sqrt{n}}}$ and $x_2=\ldots=x_n=\sqrt{\frac{1}{b(a+b)}}=\sqrt{\frac{1}{2n+2\sqrt{n}}}$.
Using a similar argument, one can also maximize $\left(\sum_{i=1}^k\,x_i\right)\left(\sum_{i=1}^n\,x_i\right)$, where $k\in\{1,2,\ldots,n\}$.  It is easy to show that $\left(\sum_{i=1}^k\,x_i\right)\left(\sum_{i=1}^n\,x_i\right)\leq \frac{\sqrt{kn}+k}{2}$ and the equality is achieved when $x_1=\ldots=x_k=\sqrt{\frac{\sqrt{n}+\sqrt{k}}{2k\sqrt{n}}}$ and $x_{k+1}=\ldots=x_n=\sqrt{\frac{1}{2\sqrt{n}(\sqrt{n}+\sqrt{k})}}$.
